Problem: Find the gradient of $f(x, y, z) = z \ln(y) + e^2$ at $(1, e, 0)$. $\nabla f(1, e, 0) = ($
Answer: The gradient of a scalar field is all its partial derivatives put together into a vector. For a 3D scalar field, this looks like $\nabla f = (f_x, f_y, f_z)$. Let's find $f_x$, $f_y$, and $f_z$. $\begin{aligned} f_x &= \dfrac{\partial}{\partial x} \left[ z \ln(y) + e^2 \right] \\ \\ &= 0 \\ \\ f_y &= \dfrac{\partial}{\partial y} \left[ z \ln(y) + e^2 \right] \\ \\ &= \dfrac{z}{y} \\ \\ f_z &= \dfrac{\partial}{\partial z} \left[ z \ln(y) + e^2 \right] \\ \\ &= \ln(y) \end{aligned}$ Now we can evaluate the partial derivatives we found at the point $(1, e, 0)$. $\begin{aligned} f_x(1, e, 0) &= 0 \\ \\ f_y(1, e, 0) &= \dfrac{z}{y} = 0 \\ \\ f_z(1, e, 0) &= \ln(y) = 1 \end{aligned}$ The gradient of $f$ at $(1, e, 0)$ is $\nabla f = (0, 0, 1)$.